void buzz(int num, int [] a)
{
for(int i=1;i<num;i*=2){
for(int j=0;j<a[i]%num;j++){
print(num);
}
}
}
I understand that the outer for loop runs for log n times. Need help in understanding the complexity of the inner loop and the overall.
Let us assume num = 2^n for simplicity.
The inner loop runs a[i] % num times, which is comprised in range [0, num-1]. The outer loop runs n times so you can conclude O(num.log(num)) = O(n.2^n) worst-case complexity.
For a precise evaluation, let r[i] := a[i] % num. Then the number of print's is
r[1] + r[2] + r[4] + r[8] + ... r[2^(n-1)].
This is n times the average r (for the relevant indexes; the expectation of r could be (num-1)/2)).
